Locating roots and iteration

Change of sign

If f(x) is continuous on [a, b] and f(a) and f(b) have opposite signs, then f(x) = 0 has a root between a and b. Always say both: 'sign change and f is continuous, so there is a root'.

It can fail: an asymptote (such as 1/(x - 2)) gives a sign change with no root; two roots close together, or a repeated root, give no sign change.

Showing a root to a given accuracy

To show a root is 2.279 to 3 decimal places, test the bounds 2.2785 and 2.2795. A sign change between them proves it.

Iteration

Rearrange f(x) = 0 into x = g(x), then use x_n+1 = g(x_n) from a starting value x_0. Some rearrangements converge to the root and some diverge.

On a graph of y = x and y = g(x): a staircase diagram approaches the root from one side; a cobweb diagram spirals in, alternating either side.

Worked example

f(x) = x³ - 3x - 5. Show there is a root between 2 and 3, then use x_n+1 = (3x_n + 5)^1/3 with x_0 = 2 to find x_1.

1. f(2) = 8 - 6 - 5 = -3 < 0 and f(3) = 27 - 9 - 5 = 13 > 0.
2. f is continuous and changes sign, so there is a root in (2, 3).
3. x_1 = (3 × 2 + 5)^1/3 = 11^1/3 = 2.2240

Answer: x_1 = 2.2240

Check yourself

1. f(x) = x³ - 3x - 5 has f(2) = -3 and f(3) = 13. What can you conclude?
   • The root is at x = 2.5, halfway between
   • There are exactly two roots between 2 and 3
   • There is no root, as f(2) is negative
   • f is continuous and changes sign, so there is a root between 2 and 3
   Show answer

   f is continuous and changes sign, so there is a root between 2 and 3. A continuous function that changes sign must cross zero.

2. f(x) = 1/(x - 2) has f(1) = -1 and f(3) = 1, but no root between 1 and 3. Why?
   • The interval is too wide
   • f is not continuous: it has an asymptote at x = 2
   • f(1) and f(3) have the same size
   • The method only works for polynomials
   Show answer

   f is not continuous: it has an asymptote at x = 2. The sign change happens across the asymptote, not at a root.

3. Use x_n+1 = (3x_n + 5)^1/3 with x_0 = 2 to find x_1, to 4 decimal places.
   Show answer

   2.2240. x_1 = 11^1/3 = 2.2240.

4. Continue the iteration x_n+1 = (3x_n + 5)^1/3, x_0 = 2, to find x_3, to 3 decimal places.
   Show answer

   2.277. x_1 = 2.2240, x_2 = 2.2684, x_3 = 2.2770.

5. Which is a correct rearrangement of x³ - 3x - 5 = 0?
   • x = (3x - 5)^1/3
   • x = (x³ + 5)/3
   • x = (3x + 5)^1/3
   • x = (5 - 3x)^1/3
   Show answer

   x = (3x + 5)^1/3. x³ = 3x + 5, then take the cube root.

6. Which pair of values should you test to show that a root of x³ - 3x - 5 = 0 is 2.279 to 3 decimal places?
   • 2.2785 and 2.2795
   • 2.278 and 2.280
   • 2.27 and 2.28
   • 2.279 and 2.280
   Show answer

   2.2785 and 2.2795. Values that round to 2.279 lie between 2.2785 and 2.2795.

7. On a cobweb diagram for a converging iteration, how do the iterates behave?
   • They approach the root from one side only
   • They move steadily away from the root
   • They alternate either side of the root
   • They all equal x_0
   Show answer

   They alternate either side of the root. A cobweb spirals in towards the root, crossing from one side to the other each step.

8. Use x_n+1 = 2 + ln x_n with x_0 = 3 to find x_2, to 3 decimal places.
   Show answer

   3.131. x_1 = 2 + ln 3 = 3.0986, x_2 = 2 + ln 3.0986 = 3.131.

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Sources

• Pearson Edexcel International Advanced Level in Mathematics specification (issue 3)
• Pearson Edexcel IAL Pure Mathematics 3 scheme of work

Last updated 7 November 2026. Spotted a mistake? Tell us.